Chapter 4 A First Analysis of Feedback Feedback Control A Feedback Control seeks to bring the measured quantity to its desired value or set-point (also called reference trajectory) automatically. There are two possible causes of difference between measured value and desired value: Disturbance and noise Change of set point, where the control system must act to bring the measured quantity to the new set point

The Three-Term Controller: PID Control Chapter 4 A First Analysis of Feedback The Three-Term Controller: PID Control The PID Controller is the most popular feedback control algorithm used in process control and industries. It is robust yet simple, easy to understand, and can provide excellent control performance despite the variation of dynamic characteristics of the process. As the name suggests, the PID Controller consists of three basic terms: the Proportional term, the Integral term, and the Derivative term. Each term can be activated separately (P, I, or D) or combined to obtain the desired controller. Depending on the characteristics of the process, the following controller are generally used: P, PI, PD, or PID.

The Three-Term Controller: PID Control Chapter 4 A First Analysis of Feedback The Three-Term Controller: PID Control Provides excitation to the process The control variable or manipulated variable To be controlled Measured variable The input to the controller, the tracking error, will vary due to the change in measured value or desired value. In response, the controller produces control signal in order to minimize the error.

first step to close a feedback loop Chapter 4 A First Analysis of Feedback Proportional Term P – Term Proportional Gain, first step to close a feedback loop

Chapter 4 A First Analysis of Feedback Integral Term I – Term Integral Gain, to assure zero error to constant reference and disturbance inputs How if P and I are combined together?

Derivative Term D – Term Chapter 4 A First Analysis of Feedback Derivative Term D – Term Derivative Gain, to improve (or even realize) stability and good dynamic response How if P and D are combined together? How if P, I, and D?

PID Control Altogether, the PID Controller may be represented as: Chapter 4 A First Analysis of Feedback PID Control Altogether, the PID Controller may be represented as: Another way to formulate PID Controller in time domain is: Integral time constant Derivative time constant

Chapter 4 A First Analysis of Feedback Effects of Each Term The proportional term makes the response of the system faster and reduces error due to disturbance by increasing the overall gain of the system, but it alone may still allow a steady-state error. The integral term may eliminate the steady-state error to a constant input, although at the cost of deteriorating the dynamic response of the system by causing slower response of the system to varying set point. The derivative term typically makes the system faster and better damped, although at the cost of less stability and of high sensitivity to high frequency disturbance and sensor noise. Note that the effects of each term may not be accurately identifiable because P, I, and D gains are dependent of each other

Problem Example Consider the mass-spring-damper system below. Chapter 4 A First Analysis of Feedback Problem Example Consider the mass-spring-damper system below. The dynamic model of the process can be derived as: Taking the Laplace Transform yields: The transfer function is then given by:

Chapter 4 A First Analysis of Feedback Problem Example Let m = 1 kg, b = 10 N.s/m, and k = 20 N/m, while the applied force F = 1 N. The transfer function can now be written as: The process will now be controlled by a PID-Controller. The contribution of each controller parameter kp, ki, and kd will be observed on an effort to obtain an overall system with small rise time, minimum overshoot, small settling time, and zero steady-state error.

Problem Example: Open-Loop Chapter 4 A First Analysis of Feedback Problem Example: Open-Loop The open-loop transfer function of the process is: The steady-state output value of the system is: The gain of the process is 0.05

Problem Example: Open-Loop Chapter 4 A First Analysis of Feedback Problem Example: Open-Loop The open-loop unit-step response of the process can be shown as: No overshoot Settling time is around 1.5 s

Problem Example: Closed-Loop Chapter 4 A First Analysis of Feedback Problem Example: Closed-Loop The process is now placed into a closed-loop system as follows: ≡ Now, the gain of the process is 1/21 = 0.0476 (decrease)

Problem Example: Closed-Loop Chapter 4 A First Analysis of Feedback Problem Example: Closed-Loop The closed-loop unit-step response of the system can be shown as: Set point 0.0476 ess = 0.9524

Problem Example: Using P-Controller Chapter 4 A First Analysis of Feedback Problem Example: Using P-Controller With the P-Controller in the loop, the transfer function of the system is given as: The parameter kp can now be tuned to obtain the desired closed-loop response of the overall system.

Problem Example: Using P-Controller Chapter 4 A First Analysis of Feedback Problem Example: Using P-Controller Let kp = 300, the closed-loop transfer function of the system is: Final value 300/320 = 0.9375  Steady-state error reduced The unit-step response is now: Overshoot increased Effects of P-Controller Rise time reduced Settling time slightly reduced

Problem Example: Using PD-Controller Chapter 4 A First Analysis of Feedback Problem Example: Using PD-Controller With the PD-Controller in the loop, the transfer function of the system is given as: The parameter kp and kd can now be tuned to obtain the desired closed-loop response of the overall system.

Problem Example: Using PD-Controller Chapter 4 A First Analysis of Feedback Problem Example: Using PD-Controller Let kp = 300, kd = 10, the closed-loop transfer function of the system is: Final value 300/320 = 0.9375  Unchanged  Steady-state error still 0.0625 The unit-step response is now: Overshoot reduced Effects of PD-Controller Settling time reduced even more Rise time almost unchanged

Problem Example: Using PI-Controller Chapter 4 A First Analysis of Feedback Problem Example: Using PI-Controller With the PI-Controller in the loop, the transfer function of the system is given as: The parameter kp and ki can now be tuned to obtain the desired closed-loop response of the overall system.

Problem Example: Using PI-Controller Chapter 4 A First Analysis of Feedback Problem Example: Using PI-Controller Let kp = 300, ki = 160, the closed-loop transfer function of the system is: Final value 160/160 = 1  Zero steady-state error The unit-step response is now: Overshoot increased Effects of PI-Controller Settling time increased Actually, both P and I terms tend to increase overshoot and reduce rise time at the same time (double effect) Rise time almost unchanged

Problem Example: Using PID-Controller Chapter 4 A First Analysis of Feedback Problem Example: Using PID-Controller With the PID-Controller in the loop, the transfer function of the system is given as: The parameter kp, ki, and kd can now be tuned to obtain the desired closed-loop response of the overall system.

Problem Example: Using PID-Controller Chapter 4 A First Analysis of Feedback Problem Example: Using PID-Controller Let kp = 300, ki = 160, kd = 10, the closed-loop transfer function of the system is: Final value 160/160 = 1  Zero steady-state error The unit-step response is now: Overshoot significantly reduced Effects of PID-Controller Settling time unchanged Small rise time, almost unchanged

Chapter 4 A First Analysis of Feedback Problem Example To ease the comparison, all the unit-step responses are plotted on the same figure. P-Controller PD-Controller PI-Controller PID-Controller Original Process

Chapter 4 A First Analysis of Feedback Conclusions Proportional control can reduce steady-state errors, but the high gains almost always destabilize the system. Integral control provides total reduction in steady-state errors, but often makes the system less stable. Derivative control usually increases damping and improves stability, but has almost no effect on the steady-state error. These 3 kinds of control combine to form the classical PID-Controllers which provide reasonable control for most industrial processes, provided that the performance demand is not too high.

Control of 1st Order System Using P-Controller Chapter 4 A First Analysis of Feedback Control of 1st Order System Using P-Controller For a unit-step reference trajectory, the steady-state error:  In order to reduce ess, kp must be increased

Control of 1st Order System Using P-Controller Chapter 4 A First Analysis of Feedback Control of 1st Order System Using P-Controller Close-loop sensitivity At high frequency, S≈1 At low frequency, if kp increased, S decreases, T increases If kp is increased to reduce ess, then the system is less vulnerable to disturbance with low frequency, but more vulnerable to sensor noise with low frequency

Control of 1st Order System Using PI-Controller Chapter 4 A First Analysis of Feedback Control of 1st Order System Using PI-Controller

Control of 1st Order System Using PI-Controller Chapter 4 A First Analysis of Feedback Control of 1st Order System Using PI-Controller For a unit-step reference trajectory, the steady-state error: At high frequency, S≈1, T≈0 At low frequency, S≈0, T≈1 The system is very sensitive to disturbance with high frequency and sensor noise with low frequency  No steady-state error

Control of 2nd Order System Using P-Controller Chapter 4 A First Analysis of Feedback Control of 2nd Order System Using P-Controller

Control of 2nd Order System Using P-Controller Chapter 4 A First Analysis of Feedback Control of 2nd Order System Using P-Controller For R(s) = 1/s, For R(s) = 1/s2,  No steady-state error  The error ess can be reduced, but cannot equal zero

Control of 2nd Order System Using P-Controller Chapter 4 A First Analysis of Feedback Control of 2nd Order System Using P-Controller In order to reduce the steady-state error , Decrease B  ζ decreased Increase kp  ωn increased while ζ decreased In any way, ζ will decrease  maximum overshoot increases

Control of 2nd Order System Using PD-Controller Chapter 4 A First Analysis of Feedback Control of 2nd Order System Using PD-Controller

Control of 2nd Order System Using PD-Controller Chapter 4 A First Analysis of Feedback Control of 2nd Order System Using PD-Controller For R(s) = 1/s2,

Control of 2nd Order System Using PD-Controller Chapter 4 A First Analysis of Feedback Control of 2nd Order System Using PD-Controller In order to reduce the steady-state error , Decrease B  ζ decreased, but can be compensated by increasing kd Increase kp  ωn increased and ζ decreased, but can be compensated by increasing kd Steady-state response (ess) can be tuned without affecting transient response (Mp)

Chapter 4 A First Analysis of Feedback Homework 5 No.1. The performance of P- and PD-Controller to control the 2nd order system are already analyzed on the previous pages. Now, conduct further analysis for PI-Controller. Explain about its ability to reduce steady-state error in the presence of step input and ramp input. No.2. A motor used for laser surgery requires high accuracy for position and velocity response. Consider the system shown below.

Homework 5 No.2. (continued) Chapter 4 A First Analysis of Feedback Homework 5 No.2. (continued) The amplifier gain K must be adjusted so that the steady- state error for a ramp input r(t) = At (where A = 1 mm/s) is less than or equals to 0.1 mm, while a stable response is maintained. (a) Determine the value of K to meet the requirement. (b) In MATLAB Simulink, build the motor system as given previously. Now, prove the answer of (a), that ess = 0.1 mm can be achieved for that certain value of K. (c) A technician claims that he can obtain ess = 0.02 mm. What is your opinion? Is his claim possible to do or not? How? Why? Explain. Deadline: 09.10.2012, at 07:30.